A chemistry student is given 700. mL of a clear aqueous solution at 26.° C. He is told an unknown amount of a certain compound X is dissolved in the solution. The student allows the solution to cool to 26.° C. The solution remains clear. He then evaporates all of the water under vacuum. A precipitate remains. The student washes, dries and weighs the precipitate. It weighs 0.032 kg. Required:
Using only the information above, can you calculate the solubility of X in water at 22.0°C ? If you said yes, calculate it. Be sure your answer has a unit symbol and 2 significant digits.

The correct answer is - yes, 4.57 g of solute per 100 ml of solution

Explanation:

The correct answer is yes we can calculate the solubility of X in the water at 22.0°C. The salt will remain after the evaporate from the dissolved and cooled down at 26°C.

Then, the amount of solute dissolved in the 700 ml solution at 26°C is the weighed precipitate: 0.032 kg = 32 g.

Then solublity will be :

32. g solute / 700 ml solution = y / 100 ml solution

⇒ y = 32. g solute × 100 ml solution / 700 ml solution = 4.57 g.

Thus, the answer is 4.57 g of solute per 100 ml of solution.

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.